im totally blanking out on math. plz help.
2006-11-07 12:57:24 · 7 answers · asked by Adeline 2 in Science & Mathematics ➔ Mathematics
(x-2)^5 = ((x-2)^2)^2*(x-2) = (x^4-8x^3+24x^2-32x+16)(x-2) =
x^5-8x^4+24x^3-32x^2+16x-2x^4+16x^3-48x^2+64x-32 =
x^5 -10x^4 +40x^3 -80 x^2 +80x -32
2006-11-07 13:44:34 · answer #1 · answered by Anonymous · 0⤊ 0⤋
this can be done in 2 ways. One is multiplication which is a lengthy process or 2nd is using bionomial series
nth term is x^n*(-2)^(5-n) *5Cn n is from 5 to 0
expand and collect coffeifcient
term with x^5 = x^5
x^4 = -10x^4
x^3 = 40 x^3
x^2 = -80x^2
x = 80x
constat = -32
add tem and get x^5-10x^4+40x^3-80x^2+80x-32
2006-11-07 21:08:14 · answer #2 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
What is the sign between the 5 and the bracket?
2006-11-07 21:00:14 · answer #3 · answered by Question 2 · 0⤊ 0⤋
(x-2)^5=
=(x-2)(x-2)(x-2)(x-2)(x-2)
or
(x-2)^5=
=(x-2)*(x-2)*(x-2)*(x-2)*(x-2)
* is multiplication
2006-11-07 21:04:42 · answer #4 · answered by tr_z9z9 2 · 0⤊ 0⤋
i've got
x^5-10x^4+40x^3-80x^2+80x-32
it may be wrong though....
2006-11-07 21:04:56 · answer #5 · answered by Anonymous · 0⤊ 0⤋
x^5+5C1x^4(-2)+5C2x^3(-2)^2+
5C3x^2(-2)^3+5C4x(-2)^4+(-2)^5
2006-11-07 21:02:57 · answer #6 · answered by raj 7 · 0⤊ 0⤋
ask a friend, hire a tutor(?),ask parents, look it up, etc.
2006-11-07 20:59:54 · answer #7 · answered by *gasp* it's me! 3 · 0⤊ 0⤋